Phản ứng: RCOOH + NaOH -> RCOONa + H2O. Sự tăng khối lượng khi chuyển từ acid sang muối là do thay thế H bằng Na, tức là tăng 22 g cho mỗi mol acid phản ứng. Tổng số mol acid trong hỗn hợp = (khối lượng muối - khối lượng acid) / 22 = 8.4 / 22 = 0.3818... mol. This value seems unusual. Let's re-check the calculation. m_muoi = m_acid + m_tang. m_tang = 8.4g. m_tang = n_acid * (M_Na - M_H) = n_acid * (23 - 1) = n_acid * 22. n_acid = 8.4 / 22 = 0.3818... This number seems correct based on the given increase. Let's assume the question intends to provide numbers that lead to a cleaner result or expects a specific approach. Let's try setting up equations. Let x = mol HCOOH, y = mol CH3COOH. m = 46x + 60y. m_muoi = 68x + 82y. m_muoi - m = (68x + 82y) - (46x + 60y) = 22x + 22y = 22(x+y). We are given m_muoi - m = 8.4. So, 22(x+y) = 8.4 => x+y = 8.4 / 22 = 0.3818... Let's re-read the prompt. 'thu được (m + 8.4) gam hỗn hợp muối'. This means the mass increased by 8.4g. Okay, the calculation of total moles is correct. x+y = 0.3818... We need to find % khối lượng HCOOH = (46x / (46x + 60y)) * 100%. We have one equation, x+y = 0.3818..., and two unknowns x, y. We need more information. Perhaps the question intended a different mass increase or a different total mass. Let's assume there is a typo in the increase value, and it should lead to a simple total number of moles, e.g., 0.1, 0.2, etc. If the increase was 2.2g, total moles = 0.1. If the increase was 4.4g, total moles = 0.2. If the increase was 6.6g, total moles = 0.3. If the increase was 8.8g, total moles = 0.4. Let's check if 8.4g is a multiple of 2.2. 8.4 / 2.2 = 3.818... It's not. Let's assume the total mass of the mixture 'm' was given, or the total moles were given. Since the question asks for percentage composition, usually the total mass 'm' or total moles 'x+y' would be the missing piece. Let's look at the options if they were provided in a real test. Since I have to generate options, I will assume the total moles of acid is a nice number, say 0.2 mol. If x+y = 0.2, then 22*0.2 = 4.4g increase. This doesn't match 8.4g. Let's assume the increase value 8.4g is correct. Then x+y = 8.4/22. We cannot solve for x and y individually with one equation. Let's reconsider the problem. Is there any other property that distinguishes HCOOH and CH3COOH? Only HCOOH can do silver mirror reaction. Maybe that's tested in another question. This question seems purely about stoichiometry and mass increase. Let's assume there's a constraint missing or a typo in the numbers. However, I must generate a valid question based on the topic. Let's assume the question implicitly implies something about the relative amounts, or maybe there's a relationship between x and y that's not stated. Let's search for similar problems online. Okay, similar problems usually give the total mass of the mixture and the mass of the salt, or the volume/mass of NaOH used. The increase in mass method is valid for finding total moles. Let's assume the total mass of the mixture 'm' was also given, e.g., m = 18.4g. Then 46x + 60y = 18.4 and x+y = 8.4/22. This can be solved. Let's assume for the sake of creating a complete question that total moles is 0.4 (increase 8.8g). This doesn't match 8.4g. Let's assume the question meant to give the total mass of the *salts* was m+8.4. Yes, that is stated. Let's stick with n_total = 8.4/22. We cannot proceed without more information or a typo correction. Let me try to create a solvable problem with similar structure. Suppose m = 9.2g and m_muoi = 13.6g. Increase = 13.6 - 9.2 = 4.4g. Total moles = 4.4/22 = 0.2 mol. So x+y = 0.2 and 46x + 60y = 9.2. Solve the system: 46x + 60(0.2-x) = 9.2 => 46x + 12 - 60x = 9.2 => -14x = -2.8 => x = 0.2. If x = 0.2, then y = 0. This means the mixture was pure HCOOH (9.2g HCOOH = 9.2/46 = 0.2 mol). This is a valid scenario, but doesn't test a mixture. Let's try another example. Suppose m = 10.6g and m_muoi = 15.0g. Increase = 15.0 - 10.6 = 4.4g. Total moles = 0.2. x+y = 0.2, 46x + 60y = 10.6. 46x + 60(0.2-x) = 10.6 => 46x + 12 - 60x = 10.6 => -14x = -1.4 => x = 0.1. If x = 0.1, then y = 0.1. Total mass = 46*0.1 + 60*0.1 = 4.6 + 6 = 10.6g. This works! So a mixture of 0.1 mol HCOOH and 0.1 mol CH3COOH has mass 10.6g and reacts to give salts with mass increase 4.4g. The original numbers (m and 8.4) seem problematic for a typical test question unless m is also given. Let's assume the problem *intended* to give m = 18.4g. Then x+y = 8.4/22, 46x + 60y = 18.4. Still doesn't seem right. Let's assume the increase was 4.4g and total mass was 9.2g. Then x+y=0.2, 46x+60y=9.2. 46x+60(0.2-x)=9.2 => 46x+12-60x=9.2 => -14x = -2.8 => x=0.2, y=0. This is pure HCOOH. Let's assume the increase was 4.4g and total mass was 10.6g. Then x=0.1, y=0.1. %HCOOH = (46*0.1 / 10.6) * 100% = (4.6/10.6)*100% = 43.4%. This is a plausible answer. Let's use this scenario to create the question. I will rephrase the question slightly to give the total mass of the mixture.
